Circuits

Current Current = rate of change of charge

SI Unit: 1 ampere = 1 A = 1 C/s Current direction is defined as the direction of positive “charge carriers”, but actually negative electrons move and carry charge the other way.

Current problems 1) Calculate the total charge through a 2 Amp circuit in 6 seconds.

2) Calculate the total charge that passes through a circuit in 6 seconds if the current is given by the equation I = 3t+2

Kirchhoff’s Junction Rule The sum of currents entering any junction must be equal to the sum of the currents leaving that junction. → Follows from conservation of charge for a steady flow of charge. ← 1A 2A → 2A → ← 2A

3A ↑ 4A ↑

i=?

8A →

Current Density J = current density through an element of a circuit = current per unit area

Direction = direction of conventional current (opposite direction of flow of electrons)

Current Density Define: n = number of charge carriers per unit volume e = charge of charge carriers vd = drift velocity of charge carriers

If charge carriers are negative, vd and J are in opposite directions.

Resistance Resistance is a property of a conductor. Resistance is defined as:

Resistance

Symbol

Unit

R

Ω = ohm = V/A

Resistivity Resistivity is a property of a material:

ρ = resistivity E = electric field J = current density

Resistance and Resistivity Resistivity and shape determine the resistance of a conductor:

Symbol

Unit

Resistance

R

Ω

Resistivity

ρ

Ω·m

Length

l

m

Cross-sectional area

A

m2

Resistivity Problem Two conductors are made of the same material and have the same length. Conductor A is a solid wire of diameter 1.0 mm. Conductor B is a hollow tube of outer diameter 2.0 mm and inner diameter 1.0 mm. What is the resistance ratio RA/RB?

Resistivity Problem

Ohm’s Law “Ohmic” devices always follow Ohm’s Law. ■

■

Resistance is independent of the applied potential difference Example: resistors V

R = V/I = slope

I

Ohm’s Law “Non-ohmic” devices ■

■

Resistance and resistivity vary with applied potential difference Example: semiconductor devices, lamps that heat up

V

V

I

I

Power Electrical circuits transfer energy ■

from potential sources ➢

■

batteries

to other devices ➢ ➢

resistors (toasters, heaters) motors

Rate of energy transfer

Rate of energy dissipation due to a resistance

Units Units for power should be Watts: P = VI Convert units: V·A → W

Power Problem A nichrome heating wire has a resistance of 72Ω. At what rate is energy dissipated in the wire if a potential difference of 120V is applied across the wire?

Emf (= voltage = potential) Circuits need a “charge pump” to produce a steady flow of charge. “Emf devices” provide an emf E (constant potential source) Example: battery. Often use E for battery potential (voltage)

Batteries Real batteries have ■ ■

Emf E internal resistance r i→

→

+ r

i

E ←i

→

R

i

Circuits Solving circuits: Total potential difference around 1 complete loop = 0 Kirchhoff’s rule: Sum of currents in = sum of currents leaving a junction

Series Circuits Series Potential V

Vtot = V1 + V2 + V3 + …

Current I

I = I1 = I2 = I3 = …

Resistance R

Req = R1 + R2 + R3 + …

Capacitance C

Parallel Circuits Parallel Potential V

V = V 1 = V2 = V3 = …

Current I

Itot = I1 + I2 + I3 + …

Resistance R Capacitance C

Ceq = C1 + C2 + C3 + …

Simple circuit problem Around 1 loop:

R1 = 4.0 Ω

E1 = 3.0 V

R2 = 5.0 Ω

Simple circuit equations Generate an equation for the following circuit, using the loop rule (summing the potential differences around the circuit loop).

Circuit Problem 1) Name currents in the 3 different branches of the circuit. 2) Write 3 equations that can be used to solve for the unknown currents. R2

R1

R3

Circuit Problem

R2

R1

I1

R3

I3

I2

Circuit Problem Calculate the currents in all 3 branches: R3 = 4.0 Ω

R2 = 2.0 Ω

I = 5/19 A ↓

E1 = 3.0 V

R1 = 5.0 Ω

← I = 8/19 A

E2 = 1.0 V

← I = 3/19 A

Calculate the rate at which energy is dissipated in R1. P = 0.346W

Circuit Problem Calculate the currents in all 3 branches: ← I = 0.5 A R = 2.0 Ω

← I = 0.25 A R = 2.0 Ω E = 6.0 V

I = 0.25 A ↑

R = 4.0 Ω

E = 3.0 V

R = 2.0 Ω I = 0.5 A →

E = 6.0 V I = 0.25 A →

R = 2.0 Ω

Capacitors Capacitors are used to store potential energy in an electric field. When a capacitor is charged to a charge of Q, its plates (opposite sides) have equal but opposite charges of +Q and –Q.

Capacitors The charge Q and potential difference V for a capacitor are proportional: Q = CV where C = capacitance depends only on geometry (not on Q or V): A = area of one side of the capacitor d = distance between capacitor plates = 8.85x10-12 F/m = permittivity of free space Unit: 1 farad = 1 F = 1C/V Practical unit: 1 pF = 1 x 10-12 F

Dielectric A dielectric is an insulating material placed between the plates of a capacitor → increases the capacitance (farads = coulombs per volt) Dielectric constant = Κ (“Kappa”)

Dielectric Given a dielectric K, replace Єo by KЄo in all electrostatic equations. Without a dielectric:

With a dielectric:

Capacitor Problem A parallel-plate capacitor has a plate area A=115cm2 and separation d=1.24cm. A potential difference Vo=85.5V is applied. a)

What is the capacitance of the capacitor?

Capacitor Problem Same capacitor: A=115cm2 d=1.24cm

Vo=85.5V C=8.21pF

What free charge appears on the plates?

q = CVo=(8.21pF)(85.5V) = 702 pC

Capacitors in Parallel A combination of capacitors in a circuit can be replaced by an equivalent capacitor In parallel:

V1 = V 2 = V 3 = … qtotal = q1 + q2 + q3 + …

Capacitors in Series In Series

Vtotal = V1 + V2 + V3 + … q1 = q2 = q3 = …

Stored Energy Work is done to charge a capacitor → Capacitors store potential energy

Capacitors in Circuits In the following circuits with capacitance and initial charge as shown, where will the charge move (if at all) when the switches are closed? 6q

3q

6q

3q

3C

C

2C

C

Capacitors in Circuits Find the equivalent capacitances of each of the following circuits. All capacitors have equal capacitance C. V

V

V V

Capacitor circuit problem ■ ■

Redraw the circuit Find V and q for capacitors C1, C2, and C3 C2 = 2 μF 2 μF 20 V

4 μF C1 = 4 μF

3 μF C3 = 3 μF

Capacitor Circuit Problem Answers: C1 C2 C3 Total

C (μF) 4 2 3 3 C1

V (V) 5 10 10 20 C2

C3

q (μC) 20 20 30 60

Capacitor Circuit Problem Same circuit. Find the energy stored in the capacitors C1, C2, and C3: C q UC V (V) (μF) (μC) (μJ) C1

4

5

20

50

C2

2

10

20

100

C3

3

10

30

150

Total

3

20

60

600

Time Dependence When capacitors are first connected: ■ ■ ■

Initially no charge on capacitor Charge rapidly Act like a wire

VVCC

VVRR

tt

tt

Time Dependence After being connected for a long time … ■ ■ ■

Capacitor is fully charged Can not charge any more Act like an open circuit (no current)

VCVC

VVRR

tt

tt

Time Dependence When capacitors are first disconnected from the battery: ■ ■ ■

Initially full charge on capacitor Discharge rapidly Act like a battery

VVCC

tt

Time Dependence When capacitors have been disconnected for a long time: ■ ■

Towards no more charge on the capacitor Exponentially to no more current

VV CC

tt

RC Circuit With C completely uncharged, the switch is suddenly closed (at t = 0). Determine the current in the circuit at t = 0 and as t → ∞. R = 4.70 kΩ C = 1.5 pF R = 6.0 V C

RC Circuit

With C completely uncharged, the switch is suddenly closed (at t = 0). Determine the current through each resistor at t = 0 and as t → ∞. R1 = R2 = R3 = 0.73 MΩ R1 C = 6.5 μF R3 = 1.2 kV R2

C

Charging a capacitor: Around the loop:

u substitution:

… charging a capacitor

RC Circuits: Charging Charge: Current: Potential:

Time constant:

RC Circuits: Discharging Charge: Current: Potential:

Time constant:

Capacitor problem A capacitor of capacitance C is discharging through a resistor of resistance R. In terms of the time constant when will the charge on the capacitor be half its initial value?

Capacitors charging In reality, capacitors take some time to charge and to discharge: http://www.splung.com/content/sid/3/page/capacitors

Discharging a capacitor Around the loop: R C

This is the differential equation. It has a variable (q) and the derivative of the variable (dq/dt).

… discharging a capacitor … Separate variables:

Success: our variable (q) is on one side of the equation, and time and constants are on the other side of the equation.

… discharging a capacitor ... Integrate:

… discharging a capacitor.

Solve for our variable (q) and make it pretty:

Inductors Inductors are little solenoids. An induced emf appears in any coil (an inductor) in which the current is changing. Self-induced emf

L = inductance of inductor unit = henry = H = T·m2/A L is a measure of the opposition to the rate of change of current

Inductance Inductors oppose change in a circuit. → I increasing

→ I decreasing

Initially, an inductor acts to oppose changes in the current through it. A long time later, it acts like ordinary connecting wire.

Uses for Inductors ■

Used extensively in analog circuits and signal processing ■

■

■

for tuning frequencies, filtering out specific frequencies

2 or more inductors with coupled magnetic flux form a transformer Large, heavy components; used less in modern equipment

RL Circuits Close circuit, start current:

Open circuit, stop current:

Inductive time constant:

RL Circuit An RL circuit is hooked up, as shown. Plot the potential across the resistor and across the inductor as a function of time: VR

VL

t

t

RL Circuits What is the current through the battery just after the switch is closed? Battery has = 18V Resistors are all R = 9 Ω Inductors are all L = 2 mH

RL Circuits What is the current through the battery after a long time? Battery has = 18V Resistors are all R = 9 Ω Inductors are all L = 2 mH

RL Circuits Rank the circuits according to the current through the battery a) just after the switch is closed and b) a long time later.

Inductors Energy is stored in an inductor:

RL Circuit Problem A solenoid has an inductance of 53 mH and a resistance of 0.37 Ω. If it is connected to a battery, how long will the current take to reach half its final equilibrium value?

RL Circuit Problem Same circuit: L=53mH R=0.37Ω If the circuit is attached to a 12V battery, how much energy is stored after 0.10s (at half the equilibrium current)?

LC Circuit First charge the capacitor with the switch at a, then throw the switch to b. a

b

C

L

Describe the subsequent current in the circuit.

→ It oscillates!

LC circuit Write (but do not solve) the equation for the charge and/or current in the LC circuit when the switch is at b. a C

b L